2019版高考數(shù)學(xué)復(fù)習(xí)專(zhuān)題四數(shù)列2.4.2.2求數(shù)列的通項(xiàng)及前n項(xiàng)和課件文.pptx

1、4.2.2求數(shù)列的通項(xiàng)及前n項(xiàng)和,-2-,考向一,考向二,考向三,求數(shù)列的通項(xiàng)及錯(cuò)位相減求和,例1(2018河北唐山一模,文17)已知數(shù)列an是以1為首項(xiàng)的等差數(shù)列,數(shù)列bn是以q(q1)為公比的等比數(shù)列,且a2=b1,a3-a1=b2-b1,a2b2=b3. (1)求an和bn的通項(xiàng)公式; (2)若Sn=a1bn+a2bn-1+an-1b2+anb1,求Sn.,-3-,考向一,考向二,考向三,-4-,考向一,考向二,考向三,解題心得若已知數(shù)列為等差或等比數(shù)列,求其通項(xiàng)是利用等差、等比數(shù)列通項(xiàng)公式,或通過(guò)變形轉(zhuǎn)換成等差、等比數(shù)列求通項(xiàng);如果數(shù)列an與數(shù)列bn分別是等差數(shù)列和等比數(shù)列,那么數(shù)列a

2、nbn的前n項(xiàng)和采用錯(cuò)位相減法來(lái)求.,-5-,考向一,考向二,考向三,對(duì)點(diǎn)訓(xùn)練 1(2018山東濰坊一模,理17)公差不為0的等差數(shù)列an的前n項(xiàng)和為Sn,已知S4=10,且a1,a3,a9成等比數(shù)列. (1)求an的通項(xiàng)公式;,-6-,考向一,考向二,考向三,-7-,考向一,考向二,考向三,-8-,考向一,考向二,考向三,-9-,考向一,考向二,考向三,解題心得對(duì)于已知等式中含有an,Sn的求數(shù)列通項(xiàng)的題目,一般有兩種解題思路,一是消去Sn得到f(an)=0,求出an;二是消去an得到g(Sn)=0,求出Sn,再求an. 把數(shù)列的通項(xiàng)拆成兩項(xiàng)之差,求和時(shí)中間的項(xiàng)能夠抵消,從而求得其和.注意抵

3、消后所剩余的項(xiàng)一般前后對(duì)稱(chēng).,-10-,考向一,考向二,考向三,對(duì)點(diǎn)訓(xùn)練2(2018河北石家莊一模,文17)已知an是公差不為零的等差數(shù)列,滿(mǎn)足a3=7,且a2,a4,a9成等比數(shù)列. (1)求數(shù)列an的通項(xiàng)公式; (2)設(shè)數(shù)列bn滿(mǎn)足bn=anan+1,求數(shù)列 的前n項(xiàng)和Sn.,-11-,考向一,考向二,考向三,-12-,考向一,考向二,考向三,求數(shù)列的通項(xiàng)及分項(xiàng)求和 例3(2018山東濟(jì)寧一模,理17)已知an是等比數(shù)列,滿(mǎn)足a1=2,且a2,a3+2,a4成等差數(shù)列,數(shù)列bn滿(mǎn)足 (1)求an和bn的通項(xiàng)公式; (2)設(shè)cn=(-1)n(an-bn),求數(shù)列cn的前2n項(xiàng)和S2n.,-1

4、3-,考向一,考向二,考向三,解: (1)設(shè)數(shù)列an的公比為q,則由條件得2(a3+2)=a2+a4.又a1=2, 則2(2q2+2)=2q+2q32(q2+1)=q(1+q2). 1+q20,q=2,故an=2n. 對(duì)于bn,當(dāng)n=1時(shí),b1=21=2;,-14-,考向一,考向二,考向三,解題心得若能把一個(gè)數(shù)列的通項(xiàng)分成一部分是等差數(shù)列通項(xiàng),另一部分是等比數(shù)列,則其前n項(xiàng)和分成了兩個(gè)數(shù)列的前n項(xiàng)和,分別求和后相加即可;同理,若一個(gè)數(shù)列的前n項(xiàng)和不好求,對(duì)其通項(xiàng)變形后,如果能分成兩個(gè)部分,每一部分的前n項(xiàng)和能求,則問(wèn)題得到解決.,-15-,考向一,考向二,考向三,對(duì)點(diǎn)訓(xùn)練 3(2018福建龍巖4月質(zhì)檢,文17)已知正項(xiàng)等比數(shù)列an的前n項(xiàng)和為Sn,且Sn=2an-1(nN*). (1)求數(shù)列an的通項(xiàng)公式; (2)若bn=lg an,求數(shù)列an+bn的前n項(xiàng)和Tn.,-16-,考向一,考向二,考向三,解: (1)由Sn=2an-1(nN),可得S1=2a1-1, a1=2a1-1.a1=1. S2=2a2-1,a1+a2=2a2-1, a2=2.數(shù)列an是等比數(shù)列, 數(shù)列an的通項(xiàng)公式為an=2n-1. (2)由(1)知,bn=lg an=(n-1)l